Lecture 3
Dr. Sarah M.Eljack
In Propositional Logic, we assume a collection
of atomic propositions are given: p, q, r, s, t,
….
Then we form compound propositions by using
logical connectives (logical operators) to form
propositional “molecules”.
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2
Operator
Negation
Conjunction
Disjunction
Exclusive or
Conditional
Biconditional
Symbol Usage






Java
not
!
and
&&
or
||
xor
(p||q)&&(!p||!q)
if,then
p?q:true
iff
(p&&q)||(!p&&!q)
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p = “Cruise ships only go on big rivers.”
q = “Cruise ships go on the Hudson.”
r = “The Hudson is a big river.”
r = “The Hudson is not a big river.”
pq = “Cruise ships only go on big rivers and
go on the Hudson.”
p q
r = “If cruise ships only go on big
rivers and go on the Hudson, then the
Hudson is a big river.”
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This just turns a false proposition to true and
the opposite for a true proposition.
EG: p = “23 = 15 +7”
p happens to be false, so p is true.
In Java, “!” plays the same role:
!(23 == 15+7)
has the boolean value true whenever
evaluated.
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Logical operators are defined by truth tables –
tables which give the output of the operator
in the right-most column.
Here is the truth table for negation:
p
F
T
p
T
F
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Conjunction is a binary operator in that it
operates on two propositions when creating
compound proposition. On the other hand,
negation is a unary operator (the only nontrivial one possible).
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Conjunction is supposed to encapsulate what
happens when we use the word “and” in
English. I.e., for “p and q ” to be true, it must
be the case that BOTH p is true, as well as q.
If one of these is false, than the compound
statement is false as well.
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EG. p = “Clinton was the president.”
q = “Monica was the president.”
r = “The meaning of is is important.”
Assuming p and r are true, while q false.
Out of pq, pr, qr
only pr is true.
Java: x==3 && x!=3
Evaluates to false for any possible value of x.
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p
q
p q
T
T
F
F
T
F
T
F
T
F
F
F
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Conversely, disjunction is true when at least
one of the components is true:
p
q
p q
T
T
F
F
T
F
T
F
T
T
T
F
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Note: English version of disjunction “or” does
not always satisfy the assumption that one of
p/q being true implies that “p or q ” is true.
Q: Can someone come up with an example?
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A: The entrée is served with soup or salad.
Most restaurants definitely don’t allow you to
get both soup and salad so that the
statement is false when both soup and salad
is served. To address this situation,
exclusive-or is introduced next.
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p
q
p q
T
T
F
F
T
F
T
F
F
T
T
F
Note: in this course any usage of “or”
will connote the logical operator 
as opposed to the exclusive-or.
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This one is probably the least intuitive. It’s
only partly akin to the English usage of
“if,then” or “implies”.
DEF: p  q is true if q is true, or if p is false.
In the final case (p is true while q is false) p 
q is false.
Semantics: “p implies q ” is true if one can
mathematically derive q from p.
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p
q
p q
T
T
F
F
T
F
T
F
T
F
T
T
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Q: Does this makes sense? Let’s try
examples for each row of truth table:
1.
If birds can fly then birds can fly.
2.
If birds can fly then birds like meats
3.
If birds like meats then birds can fly.
4.
If birds like meats then birds like meats
We make the assumption that “birds can fly”
is a true propositions, while “birds like
meats ” is a false propositions.
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A:
1.
2.
3.
4.
If birds can fly then birds can fly.
True: nothing about this statement is false.
If birds can fly then birds like meats
False: seems to assert falsehood
If birds like meats then birds can fly.
True: argument for –only care about end-result.
Argument against –counters common English
hyperbole.
If birds like meats then birds like meats
True. WAIT! By first reasoning in 3, when “if” part is
false, should only care about “then” part!!!!!
On other hand, standard English hyperbole.
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Remember, all of these are mathematical
constructs, not attempts to mimic English.
Mathematically, p should imply q whenever
it is possible to derive q by from p by using
valid arguments. For example consider the
mathematical analog of no. 4:
If 0 = 1 then 3 = 9.
Q: Is this true mathematically?
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A: YES mathematically and YES by the truth
table.
Here’s a mathematical proof:
1. 0 = 1 (assumption)
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A: YES mathematically and YES by the truth
table.
Here’s a mathematical proof:
1. 0 = 1 (assumption)
2. 1 = 2 (added 1 to both sides)
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A: YES mathematically and YES by the truth
table.
Here’s a mathematical proof:
1. 0 = 1 (assumption)
2. 1 = 2 (added 1 to both sides)
3. 3 = 6 (multiplied both sides by 3)
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A: YES mathematically and YES by the truth
table.
Here’s a mathematical proof:
1.
2.
3.
4.
0
1
3
0
=
=
=
=
1
2
6
3
(assumption)
(added 1 to both sides)
(multiplied both sides by 3)
(multiplied no. 1 by 3)
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A: YES mathematically and YES by the truth
table.
Here’s a mathematical proof:
1.
2.
3.
4.
5.
0
1
3
0
3
=
=
=
=
=
1
2
6
3
9
(assumption)
(added 1 to both sides)
(multiplied both sides by 3)
(multiplied no. 1 by 3)
(added no. 3 and no. 4)
QED
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As we want the conditional to make sense in the
semantic context of mathematics, we better
define it as we have!
Other questionable rows of the truth table can
also be justified in a similar manner.
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There are many ways to express the
conditional statement p  q :
If p then q. p implies q. If p, q.
p only if q. p is sufficient for q.
Some of the ways reverse the order of p
and q but have the same connotation:
q if p. q whenever p. q is necessary for p.
To aid in remembering these, I suggest
inserting “is true” after every variable:
EG: “p is true only if q is true”
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For p  q to be true, p and q must have
the same truth value. Else, p
 q is false:
p
q
pq
T
T
F
F
T
F
T
F
T
F
F
T
Q : Which operator is
 the opposite of?
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A:
 has exactly the opposite truth table
as .
This means that we could have defined the
bi-conditional in terms of other previously
defined symbols, so it is redundant. In fact,
only really need negation and disjunction to
define everything else.
Extra operators are for convenience.
Q: Could we define all other logical operations
using only negation and exclusive or?
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A: No. Notice that negation and exclusive-or
each maintain parity between truth and false:
No matter what combination of these
symbols, impossible to get a truth table with
four output rows consisting of 3 T’s and 1 F
(such as implication and disjuction).
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Electronic computers achieve their
calculations inside semiconducting
materials. For reliability, only two stable
voltage states are used and so the most
fundamental operations are carried out by
switching voltages between these two stable
states.
In logic, only two truth values are allowed.
Thus propositional logic is ideal for
modeling computers. High voltage values
are modeled by True, which for brevity we
call the number 1, while low voltage values
are modeled by False or 0.
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Thus voltage memory stored in a computer
can be represented by a sequence of 0’s
and 1’s such as
01 1011 0010 1001
Another portion of the memory might look
like
10 0010 1111 1001
Each of the number in the sequence is called
a bit, and the whole sequence of bits is
called a bit string.
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It turns out that the analogs of the logical
operations can be carried out quite easily
inside the computer, one bit at a time. This
can then be transferred to whole bit strings.
For example, the exclusive-or of the
previous bit strings is:
01 1011 0010 1001
 10 0010 1111 1001
11 1001 1101 0000
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Worked out on the black-board.
1.1.6.c, 1.1.22.d, 1.1.39:
1.
q = “You miss the final exam.”
r = “You pass the course.”
Express q  r in English.
2. Construct a truth table for p q.
3. Can one determine relative salaries of F
(Fannie), J (Janice) and M (Maggie) from the
following?
1. If F is not highest paid, then J is.
2. If J is not lowest paid, then M is highest paid.
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 Thank
 Any
you
questions ???